Answer
$v=0.91c=2.7\times10^8m/s$
Work Step by Step
The proper time between heartbeats is $\Delta t_o=1.0 s$, or 60 beats per minute. Because of the astronaut’s motion, this interval is dilated to $\Delta t=2.4 s$ (i.e, 25 beats per minute).
Solve for the speed from the time dilation relationship, equation 26–1a.
$$\Delta t_o = \Delta t \sqrt{1-\frac{ v^{2} } { c^{2} }} $$
$$v=c\sqrt{1-(\frac{\Delta t_o}{\Delta t})^2}$$
$$v=c\sqrt{1-(\frac{1.0s}{2.4s })^2}$$
$$v=0.91c=2.7\times10^8m/s$$